C_(r+1)=t,a_(r+1)=(t+1)(s-1)序为(s,t)的距离正则图

Distance Regular Graph for the Order of (s,t) of C_(r++)=t,a_(r+1)=(t+1)(s-1)

设Γ是序为(s,t)直径为d的距离正则图,讨论了l(c,a,b)表示在交叉阵列t(Γ)中列(c,a,b)的个数,记r=r(Γ)=l(c_1,a_1,b_1),s′=s′(Γ)=l(c_(r+1),a_(r+1),b_(r+1)),t′=t′(Γ)=l(c_(r+s′+1),a_(r+s′+1),b_(r+s′+1)).所得结论如下:设Γ=(X,E)是一个序为(s,t)的直径为d的距离正则图,如果c_(r+1)=t,a_(r+1)=(t+1)(s-1),则d=r+t′+2.

Suppose Г to be a distance regular graph with the diameter of d as the number of the intersection diagram of l（c,a,b）is discussed in the line of （c,a, b）,If r=r（Г）=1（c1,a1,b1）= s＇=s＇（Г） =1（c（r＋1） ,a（r＋1）,b（r＋1） ,t＇ = t＇（Г） = 1（c（r＋1）,a（r＋1）,b（r＋1））, the result is： Suppose Г=（X,E） is a distance regular graph with the diameter of d and an order of （s, t）,if c（r＋1） = t,a（r＋1）=（t＋1）（s-1）,the result is d=r＋t＇＋2.